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MrRa [10]
3 years ago
5

What would decrease the resistance of wires carrying an electric current

Physics
2 answers:
Anettt [7]3 years ago
6 0
The answer is A. shorter wires

rewona [7]3 years ago
4 0
Make the wires shorter, thicker, and cooler.
Also, use silver wires. If those are too expensive, then at least make sure you're using copper wires.
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Two blocks are connected by a massless rope. The rope passes over an ideal (frictionless and massless) pulley such that one bloc
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Explanation:

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3 years ago
Around what year can Carbon-14 dating be used up to and why?
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Idk<span>Around what year can Carbon-14 dating be used up to and why?</span>
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A 85 kg skier races down a hill from a height of 100 m. Ignore friction.
Ede4ka [16]

Answer:

(a)158.9 N up the ski slope

(b) 0.19

Explanation:

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F^k=MkN

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3 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charg
bekas [8.4K]

Answer:

5.3\times 10^3 N/C

Explanation:

We are given that

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1 cm=10^{-2} m

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Using 1 nC=10^{-9} C

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

E=\frac{\sigma}{2\epsilon_0}

E_1=\frac{\sigma}{2\epsilon_0}

E_2=\frac{\sigma}{2\epsilon_0}

E=E_1+E_2

E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Where \epsilon_0=8.85\times 10^{-12}C^2/Nm^2

Substitute the values

E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}

E=5.3\times 10^3 N/C

Hence, the magnitude of E in the region between the plates=5.3\times 10^3 N/C

5 0
3 years ago
A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s
Jlenok [28]

Answer:

The applied torque is 3.84 N-m.      

Explanation:

Given that,

Moment of inertia of the wheel is 2\ kg-m^2

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Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m

So, the applied torque is 3.84 N-m.

4 0
3 years ago
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